Question

If alpha and beta are the zeroes of the quadratic polynomial f(x)=6x^2+x-2, then find the value of alpha -beta ​

Answer 1

$\underline\mathfrak{Given:-}$

• $\: \: \: \: \: \: \: f \: {(x)} \: \: \leadsto \: \: {6x}^{2} \: + \: {x} \: - \: {2}$

$\underline\mathfrak{To \: \: Find:-}$

• $\: \: \: \: \: Value \: \: of \: \: {\alpha} \: - \: {\beta}?$

$\underline\mathfrak{Solutions:-}$

• $\: \: \: \: \: \: \: f \: {(x)} \: \: \leadsto \: \: {6x}^{2} \: + \: {x} \: - \: {2}$

$\: \: \: \: \: \leadsto {6x}^{2} \: + \: {x} \: - \: {2}$

$\: \: \: \: \: \leadsto {6x}^{2} \: - \: {3x} \: + \: {4x} \: - \: {2}$

$\: \: \: \: \: \leadsto {3x} \: {({2x} \: - \: {1})} \: + \: {2} \: {({2x} \: - \: {1})}$

$\: \: \: \: \: \leadsto {({3x} \: + \: {2})} \: \: \: {({2x} \: - \: {1})}$

• $\: \: \: \: \: {x} \: = \: \frac{-2}{3} \: \: \: Or \: \: \: {x} \: = \: \frac{1}{2}$

$\: \: \: \: \: \leadsto {\alpha} \: = \: \frac{-2}{3} \: \: \: Or \: \: \: {\beta} \: = \: \frac{1}{2}$

• $\: \: \: \: \: Now, \: \: putting \: \: the \: \: value \: \: of \: \: \alpha \: \: and \: \: \beta.$

$\: \: \: \: \: \: \: \fbox{{\alpha} \: - \: {\beta}}$

$\: \: \: \: \: \: \: \large\leadsto \: \: \frac{-2}{3} \: - \: \frac{1}{2}$

$\: \: \: \: \: \: \: \leadsto \: \: \frac{{-4} \: - \: {3}}{6}$

$\: \: \: \: \: \: \: \leadsto \: \: \frac{-7}{6}$

• $\: \: \: \: \: Hence, \: \: the \: \: the \: \: value \: + \: {\alpha} \: - \: {\beta} \: \: is \: \: \frac{-7}{6}$

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Answer 2

Given ,

• The α and β are the zeroes of the quadratic polynomial

• The quadratic polynomial is 6(x)² + x - 2

By middle term splitting method ,

6(x)² + 3x - 4x - 2 = 0

3x(2x + 1) - 2(2x + 1) = 0

(3x - 2)(2x + 1) = 0

x = 2/3 or x = -1/2

Therefore , the value of α and β are 2/3 and -1/2

Now , we have to find the value of α - b

Thus ,

α - β = 2/3 - (-1/2)

α - β = (4 + 3)/6

α - β = 7/6

Hence , the required value is 7/6